Why is the current the same at points A and B in a resistor?

[skepticwulf]

In an explanation of a textbook's diagram it says : "Conservation of charge requires that whatever charge flows into the resistor at point A, an equal amount of charge emerges at point B. Charge or current does not get “used up” by a resistor. So the current is the same at A and B. "
I can understand the conservation of charge but isn't current = charge divided by time?
Wouldn't the resistor SLOW the charge move, isn't the reason we use them in the first place?
So, how come the current is the same at A and B? I'd expect at B the current would be less as resistor would limit the charge passing through itself over time t ?

 

[A.T.]

but isn't current = charge divided by time?
Wouldn't the resistor SLOW the charge move

What do you mean by "SLOW"? Are you confusing distance by time with charge by time?

 

[skepticwulf]

I mean charge by time. Wouldn't resistor change the amount of charge transferred in seconds?
"So the current is the same at A and B" means charge per time is equal at each point, doesn't it? How could that be? What's a resistor for then?

 

[A.T.]

Wouldn't resistor change the amount of charge transferred in seconds?

Change compared to what? To without the resistor, yes. But the current must be the same at both ends of the resistor, because charge cannot accumulate in the resistor.

What's a resistor for then?

It resists the flow, so you need more voltage for the same current.

 

[skepticwulf]

But the voltage is not the same. If it was why would electrons flow from A to B?
At B voltage is lower, R is the same how come I-current is not changed??
Am I missing something here?

 

[A.T.]

t B voltage is lower, R is the same how come I-current is not changed?

Voltage and current are different things. Maybe this will help you:
http://en.wikipedia.org/wiki/Hydraulic_analogy#Hydraulic_analogy_with_horizontal_water_flow

 

[skepticwulf]

I know Voltage and current are different things but I just don't get why current at B is same with A, sorry maybe I'm too thick.
I need to read similar topics and do a further search and then go on.
Thanks anyway.

 

[jtbell]

To make a crude analogy, consider a stream of cars on a highway, moving at 100 km/h. As they enter a town, they slow down to 50 km/h. As they leave, they speed up again to 100 km/h. None of the cars "vanish" inside the town.

 

[phinds]

Another analogy, that I like better than jtbell's, and I'll explain why, is that current is exactly like a bicycle chain. It has to move at exactly the same rate at all points in a simple circuit loop. Think of it as the power source pulling on the bottom end of the chain, forcing it to circulate evenly all around the circuit.

The reason I like this better than jtbell's analogy (not at all that his is wrong in any way) is that his might be mistakenly interpreted as implying that the cars outside the town are still going 100mph while the ones in the town are going 50mph. This is not possible.

 

[A.T.]

The reason I like this better than jtbell's analogy (not at all that his is wrong in any way) is that his might be mistakenly interpreted as implying that the cars outside the town are still going 100mph while the ones in the town are going 50mph. This is not possible.

Of course it's possible. The cars just need to reduce their distances by half to keep the same car / time rate constant.

 

[OmCheeto]

Another analogy, that I like better than jtbell's, and I'll explain why, is that current is exactly like a bicycle chain. It has to move at exactly the same rate at all points in a simple circuit loop. Think of it as the power source pulling on the bottom end of the chain, forcing it to circulate evenly all around the circuit.

The reason I like this better than jtbell's analogy (not at all that his is wrong in any way) is that his might be mistakenly interpreted as implying that the cars outside the town are still going 100mph while the ones in the town are going 50mph. This is not possible.

I was going to agree with you, about this not being possible, but then I looked at who made the statement, and scratched my head for a couple of minutes.
It is in fact possible.
Current is a "count" of electrons per unit time.
Although the speed of the individual cars traveling through town may be lower than on the highway, there are more paths, and hence, the number of cars entering, passing through, and exiting the town would be the same.

Perhaps we need fourth analogy! "Ticket counters at a football stadium"
Now, the ticket counters would be the opposite of resistance, as each counter can admit one fan per second.
So the flow of fans into the stadium would be the reciprocal of the number of counters.
But the flow of fans before and after the counters, would be the same.

 

[phinds]

Of course it's possible. The cars just need to reduce their distances by half to keep the same car / time rate constant.

But what happens as the first car entering town slows down and the one leaving town at that time doesn't slow down. Does that not cause a problem? I'm not getting how that's possible.

 

[phinds]

Although the speed of the individual cars traveling through town may be lower than on the highway, there are more paths.

Yes, if you change the paths, I agree. That's like having some of the current go through each of 2 parallel paths in an otherwise serial loop, which causes no problem but is getting away from my bike chain analogy.

 

[Delta2]

In an explanation of a textbook's diagram it says : "Conservation of charge requires that whatever charge flows into the resistor at point A, an equal amount of charge emerges at point B. Charge or current does not get “used up” by a resistor. So the current is the same at A and B. "
I can understand the conservation of charge but isn't current = charge divided by time?
Wouldn't the resistor SLOW the charge move, isn't the reason we use them in the first place?
So, how come the current is the same at A and B? I'd expect at B the current would be less as resistor would limit the charge passing through itself over time t ?

Because there is potential difference between points A and B, the work of electric field is not zero between these two points and it would accelerate the electrons but the resistor keeps the velocity about the same by converting the work of electric field to heat (via collisions of electrons with the conductor atoms).

 

[russ_watters]

I prefer a water analogy because the above analogies each only discuss one flow scenario, whereas the OP appears confused about the relationship between two different ones.

Say you have a large tank with a small pipe coning out the bottom, with a valve in it that is all the way open. This is like a wire with no resistors. The flow rate (x GPM) is the same in the entire pipe.

Now close the valve most of the way. This is like installing a resistor in the wire. Now the flow is lower (perhaps 0.5x GPM), but as before it is the same everywhere in the pipe.

 

[A.T.]

But what happens as the first car entering town slows down and the one leaving town at that time doesn't slow down.

When the first car enters the town, how can one already be leaving it? If they drive in a continuously filled looping convoy, which one is the first?

Does that not cause a problem?

I see no problem.

 

[phinds]

When the first car enters the town, how can one already be leaving it? If they drive in a continuously filled looping convoy, which one is the first?

I see no problem.

Again, I don't follow you. Let's say there are 100 cars around the loop. 10 of them are in the town. The 10 in the town all slow down. How do the other 90 not have to slow down as well.

Alternatively, the car next in line to leave town slows down and then the one behind it slows down, and so forth, and as each car leaves town, it speed back up. Doesn't this scenario also cause a problem?

EDIT: sounds to me like maybe I'm taking my bicycle analogy too literally. Is that the case?

 

[jtbell]

Suppose you have a line of cars all going down the highway at the same speed. None of them has yet reached the edge of town, where they all have to slow down.

The first car enters town and slows down. The second car is still going at its original speed, so it gets closer and closer to the first car until it also reaches the edge of town and slows down. Now the first and second cars are closer than they were originally, and the third car has started to come closer and closer to the second car. Etc.

A similar process happens at the other edge of town, but now the cars get further apart as they successively increase their speeds.

 

[phinds]

Suppose you have a line of cars all going down the highway at the same speed. None of them has yet reached the edge of town, where they all have to slow down.

The first car enters town and slows down. The second car is still going at its original speed, so it gets closer and closer to the first car until it also reaches the edge of town and slows down. Now the first and second cars are closer than they were originally, and the third car has started to come closer and closer to the second car. Etc.

A similar process happens at the other edge of town, but now the cars get further apart as they successively increase their speeds.

And how is this not a change in the cars per unit time at a given point in the road just inside town. If I'm following this analogy right, the cars per unit time correspond to charge flow per unit time, which is current. So now, in a serial circuit you have differing currents in different parts of the circuit. Am I misreading the analogy?

 

[jtbell]

The rate at which cars pass a given point equals their linear density along the road times their speed. In units, (cars/hr) = (cars/km) · (km/hr). Inside town, if the speed is half what it is outside town, the density is double what it is outside town, that is, the cars are half as far apart inside town as they are outside of it.

 

[A.T.]

Doesn't this scenario also cause a problem?

What problem?

cars per time = local car speed / local car distance

You can vary the local speed and distance in reciprocal fashion while keeping cars per time constant.

 

[NascentOxygen]

In an explanation of a textbook's diagram it says : "Conservation of charge requires that whatever charge flows into the resistor at point A, an equal amount of charge emerges at point B. Charge or current does not get “used up” by a resistor. So the current is the same at A and B. "
I can understand the conservation of charge but isn't current = charge divided by time?
Wouldn't the resistor SLOW the charge move, isn't the reason we use them in the first place?
So, how come the current is the same at A and B? I'd expect at B the current would be less as resistor would limit the charge passing through itself over time t ?

Others have explained this.

Where the concept of "slowing down" can appear is in the description of charging a capacitor via that resistor. Because the resistor restricts the number of electrons per second, it in turn restricts the haste with which the capacitor voltage rises---the resistor in this circuit slows the capacitor's charging (and it achieves this without reducing the speed of the moving charges involved).

 

[Delta2]

I think eventually the thread starter will have to read some things about the Drude model to understand why the slowing down of electrons happens in such a way as to keep the current almost the same throughout the region of resistor.

here is the most simple page i could find from google about drude model
http://www.doitpoms.ac.uk/tlplib/thermal_electrical/drude.php

but you might also want to check wikipedia entry
http://en.wikipedia.org/wiki/Drude_model

 

[skepticwulf]

Thank you for all the replies.
What I'm stuck at is this:
Suppose 5 Columb of charge pass at A in 1 sec. The current is 5 Ampere.
Inside the resistor electrons will "slow down", they will not be able to pass the resistor like they did before A inside the conductor.
Now reaching B, less electrons will pass in 1 sec. Number of electrons passing B will be reduced. That's like squeezing the garden hose. So it is no longer 5 coloumbs per second but less, say it's 4.
So the current is now 4 Ampere.
So the current has reduced to 4 from 5.
What's wrong with this picture, that's what I'm trying to comprehend.

 

[NascentOxygen]

That's like squeezing the garden hose.

Are you saying more water flows through the first section of your garden hose than that part of it on the driveway where the car wheel is squashing it half closed?

 

[Drakkith]

Thank you for all the replies.
What I'm stuck at is this:
Suppose 5 Columb of charge pass at A in 1 sec. The current is 5 Ampere.
Inside the resistor electrons will "slow down", they will not be able to pass the resistor like they did before A inside the conductor.
Now reaching B, less electrons will pass in 1 sec. Number of electrons passing B will be reduced. That's like squeezing the garden hose. So it is no longer 5 coloumbs per second but less, say it's 4.
So the current is now 4 Ampere.
So the current has reduced to 4 from 5.
What's wrong with this picture, that's what I'm trying to comprehend.

Don't try to think of current in terms of the velocity of the charges. It's not a good way to look at it and will only confuse you.

 

[A.T.]

Don't try to think of current in terms of the velocity of the charges. It's not a good way to look at it and will only confuse you.

I second that. The car analogy got too much prominence here, because it was wrongly claimed that the scenario was impossible. But while being possible, it's is probably not the best analogy. The hydraulic or and the chain analogy seem better.

 

[davenn]

Thank you for all the replies.
What I'm stuck at is this:
Suppose 5 Columb of charge pass at A in 1 sec. The current is 5 Ampere.
Inside the resistor electrons will "slow down", they will not be able to pass the resistor like they did before A inside the conductor.
Now reaching B, less electrons will pass in 1 sec. Number of electrons passing B will be reduced. That's like squeezing the garden hose. So it is no longer 5 coloumbs per second but less, say it's 4.
So the current is now 4 Ampere.
So the current has reduced to 4 from 5.
What's wrong with this picture, that's what I'm trying to comprehend.

because you are thinking that the flow is reduced ONLY in the resistor, this is a common and incorrect understanding
The presence of the resistor reduces the flow in the WHOLE circuit, not just through the resistor

DAve

 

[Delta2]

Don't try to think of current in terms of the velocity of the charges. It's not a good way to look at it and will only confuse you.

I disagree here, at least for DC case is [itex]I=(nSe)v_{drift}[/itex] where the term inside parenthesis is clearly a constant (if we assume a resistor of constant cross section S). What the OP misses is that the resistor slows down the electrons in the same way through out all the resistor's region so that [itex]v_{drift}[/itex] is the same throughout the region of resistor. Just has to read abit about Drude model to get the picture.

 

[Drakkith]

I disagree here, at least for DC case is [itex]I=(nSe)v_{drift}[/itex] where the term inside parenthesis is clearly a constant (if we assume a resistor of constant cross section S). What the OP misses is that the resistor slows down the electrons in the same way through out all the resistor's region so that [itex]v_{drift}[/itex] is the same throughout the region of resistor. Just has to read abit about Drude model to get the picture.

Then you have a difference in drift velocity between the conductor and the resistor?

 

[OmCheeto]

Yes, if you change the paths, I agree. That's like having some of the current go through each of 2 parallel paths in an otherwise serial loop, which causes no problem but is getting away from my bike chain analogy.

I think this is why a lot of people don't like analogy. You have to be very careful how you explain it.
My analogy doesn't work if the voltage changes. (game time is approaching, people are pointing guns at the ticket counters, telling them to hurry up.)

I discovered this when I went back and "did the maths".

G = # of ticket counters ( called conductance, measured in siemens )
R = 1/G
V = 1 ( constant for simplicity )
I = V/R ( fans per second )

I=VG
G=I/V​

So, the # of ticket counters, conductance, is a bit more complicated than I posted earlier; "Now, the ticket counters would be the opposite of resistance, as each counter can admit one fan per second."

Each counters "conductance" is equal to fans/second(I) divided their level of fear of death(V).

Which even I can't understand.

 

[Dale]

That's like squeezing the garden hose.

Yes, it is very analogous to squeezing the garden hose. When you squeeze the garden hose the flow rate (liters per second passing any point) decreases along the entire hose. You do not have a 4 L/s flow where you squeeze and a 5 L/s flow throughout the rest of the hose, but when you squeeze you reduce the flow rate from 5 L/s to 4 L/s for the entire hose, including "upstream".

 

[Delta2]

Then you have a difference in drift velocity between the conductor and the resistor?

Yes ok it changes depending on wether the coefficient changes. Furthermore the velocity of all the free electrons inside the resistor is not the same for a given time t even at the DC case, it is just the average velocity that remains constant. So yes i guess you were right it is confusing (we are not saying it is wrong but confusing) to think of current in terms of the velocity of the electrons or the drift velocity (and what makes it even more confusing is that we have that linear relationship between the drift velocity and the current ).

Still, if the OP wants to get a good picture of the inner workings of what exactly is happening inside the resistor he should study some notes on the Drude model.

 

[nasu]

Then you have a difference in drift velocity between the conductor and the resistor?

Yes. And the drift velocity may be higher in the resistor than in the conductor. There is no reason to assume that the conduction electrons are slowing down in the resistor.
It all depends on the cross section and carrier concentration.

 

[skepticwulf]

Yes, it is very analogous to squeezing the garden hose. When you squeeze the garden hose the flow rate (liters per second passing any point) decreases along the entire hose. You do not have a 4 L/s flow where you squeeze and a 5 L/s flow throughout the rest of the hose, but when you squeeze you reduce the flow rate from 5 L/s to 4 L/s for the entire hose, including "upstream".

Now I've got it, finally, thank you! :)